Two DOF

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Two degrees of freedom

A system with two degrees of freedom requires at least

two independent co ordinates to completely specify theconfiguration of the system.

A system with two degrees of freedom will have two

natural frequencies & the vibratory motion is a

combination of two harmonics of these two natural

frequencies.

The vibratory motion corresponding to the natural

frequencies are known asPrincipal modes of vibration. If amplitude of vibration of one of the masses is unity, it

is known as Normal mode of vibration.

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Determine the equations of motion & the natural

frequencies of the two degree of freedom spring mass

system shown in fig.

1k

1m

1x

2

k

2m

x2

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For the 2 d.o.f system shown in fig, the equations of

motion for free, undamped vibrations are given by;

Which may be rearranged as,

0)(

and0)(

12222

2121111

xxkxm

xxkxkxm

0

and

0)(

122222

2212111

xkxkxm

xkxkkxm

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The solutions of the above harmonic motions may

be given by

where A, B & are arbitrary constants

Substituting for x1 & x2 in the equations of motion,

& canceling out sin(t+)we get,

These are homogeneous linear algebraicequations in A & B, whose solution is obtained by

equating the determinant of the coefficients A &

B to zero.

)sin(&)sin( 21 tBxtAx

0)(

0)(

2222

2212

1

BmkAk

BkAkkAm

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The determinant of coefficients A & B will be

Expanding the determinant gives;

The two natural frequencies of the system are

obtained by solving the above equation.

02222

2

2

121

mkk

kmkk

021

212

2

2

1

214

mm

kkmk

mkk

21

21

2

2

2

1

21

2

2

1

212

4

1

22 mm

kk

m

k

m

kk

m

k

m

kk

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Thus, the general solution of the equations of

motion is composed of two harmonic motions of

frequency 1 & 2.

The amplitude ratios are;

)sin()sin(

)sin()sin(

2221112

2221111

tBtBx

tAtAx

22

2

222

2

2121

2

2

2

12

2

122

2

1121

2

1

1

1

1

k

mk

mkk

k

B

Ak

mk

mkk

k

B

A

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Hence the general solutions finally become

)sin()sin(

)sin()sin(

222211112

2221111

tAtAx

tAtAx

The constants A1, A2, 1 & 2 can be evaluated

by the four initial conditions

)0(and)0(),0(),0( 2211 xxxx

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Prob.1 For the system discussed before,

Determine the natural frequencies , Modal

vectors & mode shapes. Also locate the nodefor each mode of vibration. Given m1=2 kg,

m2=1 kg, k1=40N/m & k2=20 N/m.

Ans: 1=3.162 rad/sec, 2=6.325 rad/sec.

22

2

11

1

11modesecondfor theratioAmplitude

15.0modefirstfor theratioAmplitude

B

A

B

A

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First Mode Second Mode

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Modal Vector & Mode shapes:

The normal modes of vibration corresponding to

1 & 2 can be expressed respectively as;

1 1

1

1 1 1

2 2

2

2 2 2

0.5( )

1.01.0

( )1.0

A Ax

B AA A

xB A

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0.5

1.0

-1.0

1.0

l

l2

l1


For first mode both modal displacements positive. Hence there is nonode.

For the second mode, the node (Point where amplitude is zero) may befound from the geometry of the figure.

i.e. the node is located in the middle of the two masses.

12

2

1

2

1

2

21 5.02111

1ll

l

l

l

l

l

ll

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Prob 2. Set up the equations of motion and determine

the natural frequencies & mode shapes of the two d.o.f

system shown in fig.

m m

l l l

l l l

xx1

2

T

T

T

Tsin TsinTsin

Tsin

Left mass Right mass

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1 1 2 2

Assuming the string tension T to be constant for for small oscillations,

( )From the fig, sin = , sin = , sin =

Equation of equilibrium for the left mass is

x x x x

l l l

Natural Frequencies of vibration :

1 1 2

1

1 1 21

2

1 2 22

mx sin sin 0

( ). mx 0

Equation of equilibrium for the right mass is

mx sin sin 0

( ).

2mx 0 ...........

x

(i)

m

T

T T

x x xi e T T

l lT

x xl l

T T

x x xi e T T

l l

2 1 2

2mx 0 ........ )

0

...(iiT T

x xl l

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2 2

1 2

1 2

2

A and B are arbitrary constants.

Now x sin( ) & x sin( )

Substituting in (i) & (ii) we get

2m 0.........( )

x sin( ) and x sin( )Let A t B

where

A t B t

T TA B iii

t

l l

2

2

and

2m 0.........( )

i.e2

m

2

T TA B ivl l

T T

l l

T T

l l

For a non trivial solution of A & B, the determinant of the

coefficients of A & B must be zero.

2

0

m

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2 2

2

2 2

2

2

2 2

Cross multiplying, we get,

2m 0 or

2m . Taking square root,

2 m Taking +ve sign,

2m m

T T

l l

T T

l l

T Tl l

T T

l l

2

2

or

3Taking ve sign, m or

T

lT

l

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2

11

21 11

2

2

2

22 22

Using equations (iii) & (iv) we get,

2

12

2

12

Substitut

T Tm

A l lT TB

ml l

T Tm

A l lT TB

ml l

Mode shapes :

2 2

1 1 2

1 2

1 1 2 2

3ing or and we get

1 11 1

T T T

ml ml mlA A

B B

and

2

11

21 11

2

2

2

22 22


2

12

2

12

Substitut

T Tm

A l lT TB

ml l

T Tm

A l lT TB

ml l

Mode shapes :

2 2

1 1 2

1 2

1 1 2 2

3ing or and we get

1 11 1

T T T

ml ml mlA A

B B

and

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1 2

1 1

1

1 1 1

2 2

2

2 2 2

The normal modes of vibration corresponding to &

can be expressed respectively as;1

( )1

1( )1

A Ax

B A

A AxB A

1 1

-1

1O

l l l

l1

First Mode

Second mode

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1 1

2 2

For the first mode, the two masses move in phase(i.e. both A and B are positive).Hence there is no node.

For the second mode, the two masses move out of phase.

(i.e. A and B are

Location of node :

11

1 1 1

opposite in sign) Hence node existssuch that

11 1 2 0.5

1

. the node is located at the middle of the two masses.

l l l ll l

l l l

i e

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Prob 3:

For the double pendulum shown in fig, obtain the

natural frequencies and mode shapes.

m

m

l

l

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m

m

l

l

1x

1x

2x2x -( )

mg

T2cos

T2

sinT1

T2

mg

cosT1

T2cos

T2

sin

sinT1

Equations of Motion :

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1

2

1 2 11 2 2 2 2

1 1 1 2 2

Let the tension in the upper string be T & that in the lower

string be T .From the free body diagrams of the masses,

sin , sin , T cos

cos T cos 2

x x xT mg

l lT T mg

Equations of Motion :

1 1 1 2 2

1 2 1

1

1 1 2

sin sin 0

2 0

30............( )

mg

mx T T

x x xmx mg mg

l l

mg mgmx x x i

l l

Equilibrium equation of top bob will be

Equilibrium equation of bottom bob will be

2 2 2

2 12

2 1 2

sin 0

i.e. 0

0............( )

mx T

x xmx mg

l

mg mgmx x x ii

l l

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1 2

2 2

1 2

2

Let x sin( ) and x sin( )

sin( ) and sin( )

substituting in the equations (i) & (ii), we get

30............( )

A t B t

x A t x B t

mg mgm A B iii

l l

mg mgA

l l

2

2

2

0............( )

To obtain the characteristic equation, equating the determinant of

the co efficients of A & B to zero,

3

0

m B iv

g gl l

g g

l l

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2

2 2

2

4

2

2

1

1

Expanding the determinant,

30

4

2 0


3

g g g

l l l

g g

l l

Solving

g

A l

gBl

Mode shapes :

2 11

2

22 22

10.414

3 0.5858

12.414

3 3 3.414

g

l

g gl l

g g

A l lg g gB

l l l

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The normal modes of vibration corresponding to1 & 2 can be

expressed respectively as;

1 1

1

1 1 1

2 2

22 2 2

0.414( )

1.0

2.414( )

1.0

A Ax

B A

A Ax

B A

0.414

1.0

-2.414

1.0

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Derive the frequency equation for the system shown in

fig. Assume the cord passing over the cylinder doesnot slip.

k

r

2k1

m2

m1

X

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1

1 1

2

2 1

For the mass m , the equilibrium equation is

0

Equation of equilibrium for the cylinder is

I ( ) 0

m x k x r

k r k r x r

Equations of motion :

2

2

1where I= , the mass moment of inertia of cylinder.

2m r

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2 2

2

1 1 1

2 2

1 1 2

)

Then x= sin( ) and = sin( )

substituting in (i) & (ii) and cancelling sin( ) through out,

( - ) 0..........( )

( ) 0...

A t B t

t

k m A k rB iii

k rA k k r I B

Assume x = Asin( t + & = Bsin ( t + )

2

1 1 1

2 2

1 1 2

2

2

.......( )

The frequency equation is obtained by equating the determinant

of the coefficients of A & B to zero.

( - )0

( )

1Simplifying & substituting I= , we get,

2

iv

k m k r

k r k k r I

m r

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Two simple pendulums are connected by a spring as

shown in fig. Determine the natural frequencies of the

pendulum.

m m

LL

k

a

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2 2

1 1 1 2

2 2 2

1 1 2

2 22 2 2 1

2

Taking moments about the hinged points, we have

For the left pendulum;

( ) 0( ) 0........( )

Similarly, for the right pendulum;

( ) 0

mL mgL kamL mgL ka ka i

mL mgL ka

mL

2 22 2 1

1 2

( ) 0........( )

sin( ) and sin( )

, we get

mgL ka ka ii

Let A t Let B t

Simplifying

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2 2 2 2

2 2 2 2

2 2 2 2

1

The non trivial solution for A & B may be obtained

by equating the determinant of coefficients of A & B to zero.

0

Taking +ve sign,

ra

mgL ka mL kaka mgL ka mL

mgL ka mL ka

g

L

2

2 2

d/sec Taking -ve sign,

2rad/sec.

g ka

L mL

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SEMI DEFINITE SYSTEMS

Sometimes if one of the roots of the characteristic

frequency equation is equal to zero, it indicates thatone of the natural frequencies of the system is zero.

Such systems are known as semi definite systems.

Practically this simply means that the system will moveas a rigid body without distortion of the springs that

connect the different masses of the system.

A vibrating system whose both ends are free will be a

semi definite system.

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Two blocks m1 and m2 are connected together by a spring of

constant K are resting on a smooth horizontal surface as

shown infig. Obtain an expression for the natural frequencies

for the system.

k

m1

m2

x x21

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1 1 1 2

2 2 2 1

( ) 0 and

( ) 0

m x k x x

m x k x x

1 1 1 2

2 2 1 2

0 and

0

m x kx kx

m x kx kx

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be given by


Substituting for x1 & x2 in the equations of motion,

& canceling out sin(t+)we get,



B to zero.

)sin(&)sin( 21

tBxtAx

2

2

1

2

( ) 0

( ) 0

k m A kB

kA k m B

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Expanding the determinant we get;

2

1

22

0k m k

k k m

2 2 2

1 2

4 2 2 21 2 1 2

2 2

1 2 1 2

2

12 2

1 2 1 2

1 22

1 2

( )( ) =0

( ) 0

( ) 0

Hence either 0 0.

( ) 0

( )

k m k m k

m m m m k k k

m m m m k

or m m m m k

k m m

m m

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2

1 1 112

1 1 1 1

1

1 1

1 22

1 2

2

2 11

Amplitude ratio in first mode

1 Substituting 0,

11

0

k(m )Put = , we get the amplitude ratio in second mode

m

k(m

A k mkB k m k

A k k

B k k

m

m

A k

Bk m

Mode Shapes & Modal Vectors :

2

1

22 1

1 2 2

1 1

1)1 1

m

m

m

m m

m m

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1 1

1 1 1

22 2

1

2 2 2

1

1

1

A A

B A

mA A

mB A

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1 1

-1

Node

First ModeSecond Mode

m2 m

1/

x

L

2

1 2

1

2 12 2

1 1

2

1

1 2

Let the node be at a distance x from the left end;

Taking ratio of corresponding sides from the two triangle,

1( )

( )

1

mm m

L x xmx L x

m mm m

mx

m m

L x xm m m

L

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For the Torsional system shown in fig,

determine the natural frequencies, modal

vectors & location of the node.2

II1

Kt

Solution: Let

and

be the angular twists in theshaft due to rotation of the rotors.

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1 1 1 2

2 2 2 1

( ) 0 and

( ) 0

t

t

I k

I k

1 1 1 2

2 2 2 1

0 and0

t t

t t

I k kI k k

Th l i f h b h i i

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be given by


Substituting for 1 & 2 in the equations of

motion, & canceling out sin(t+)we get,



B to zero.

1 2sin( ) & sin( )A t B t

2

2

1

2

( ) 0

( ) 0

t t

t t

k I A k B

k A k I B

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Expanding the determinant we get;

2

1

22

0t t

t t

k I k

k k I

2 2 2

1 2

4 2 2 21 2 1 2

2 2

1 2 1 2

2

12 2

1 2 1 2

1 22

1 2

( )( ) =0

( ) 0

( ) 0

Hence either 0 0.

( ) 0

( )

t t t

t t t

t

t

k I k I k

I I I I k k k

I I I I k

or I I I I k

k I I

I I

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2

1 1112

1 1 1 1

1

1 1

1 22

1 2

2

2

Amplitude ratio in first mode

1 Substituting 0,

11

0

k (I )Put = , we get the amplitude ratio in second mode

I

t t

t t

t

t

t

t

t

k k IAB k I k

kA

B k

I

I

kA

Bk

Mode Shapes & ModalVectors :

2

1

21 2 11

1 2 2

1 1

1k (I )1 1

It

I

I

I II

I I

M d l V t & M d h

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1 1

1 1 1

22 2

1

2 2 2

1

1

1

A A

B A

IA A

IB A

/I I

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1 1

-1

Node


/I2 I1

2

1 2

1

2 12 2

1 1

2

1

1 2

Let the node be at a distance x from the left end;

Taking ratio of corresponding sides from the two triangle,

1( )

( )

1

II I

L x xIx L x

I II I

Ix

I I

L x xI I I

L

G S

G S

G S

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Kt1

dA

dA

dB

IA

BIL

A

LB

Pinion

Gear

Rotor A

Rotor B

LA

LA LB'

dA

IA B'I

Geared SystemsActual system

Equivalent system

dA

dA

dB

IA

BIL

A

LB

Pinion

Gear

Rotor A

Rotor B

LA

LA LB'

dA

IA B'I


Equivalent system

dA

dA

dB

IA

BIL

A

LB

Pinion

Gear

Rotor A

Rotor B

LA

LA LB'

dA

IA B'I


Equivalent system

Kt1

Kt2

Kt2'

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1

The fig shows a geared system with a gear ratio = .

If the inertia of gears and shafts is negligible,

the second rotor system may be replaced by an

equivalent system B' (d =d ) such tha

A

B

B A

n

t:(i) The kinetic energy of the original system

is same as that of the equivalent system.

(ii) The strain energies of the original & equivalentsystem are the same.

(1) Equating the kinetic energies :

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2 ' '2 2

2

'

K.E of original system (B) =K.E of equivalent system (B')

1 1i.e. But '

2 2Mass moment of inertia of equivalent rotor

..

B B B B B A

BB B

A

I I

I I


2

2 ' ' 2

2

2'

2 2 ' 2

.......( )

Strain energy of section L Strain energy of section L'

1 1But ,

2 2Hence, Torsional stiffness of equivalent shaft

k

t B

B

B B

t B

Bt t

i

k k t

k k


2 2

2 2' 2 2........( )

B

B Bt t

A

k ii

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Problem

An engine drives a centrifugal pump through a2:1 speed reducer gear box. The mass momentof inertia of the engine flywheel and the pumpimpeller are 500 kg-m2 and 60 kg-m2 respectively.

The length and diameter of the engine shaft are250 mm & 50 mm respectively. The length &diameter of the pump impeller shaft are 150 mm& 40 mm respectively. The rigidity modulus of the

shaft material is 80 Gpa. Determine the naturalfrequencies and location of node.

G d S t

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dA

d

B

IA

BIL

A

LB

Pinion

Gear

Flywheel A

Impeller B

LA

LA LB'

dA

IA B'I


Equivalent system

=0.05m

=0.04m

=0.25m

=0.15m

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2 2500 , 60 ,Gear ratio n=2

Dia of engine shaft d 50 0.05

Length of engine shaft L 250 0.25

Dia of engine shaft d 40 0.04

Length of engine shaft L 150

A B

A

A

B

B

I kgm I kgm

mm m

mm m

mm m

mm

4 47 4

1

4 47 4

2

0.15

Polar Moment of inertia of engine shaft

0.05J 6.136 10

32 32Polar Moment of inertia of pump shaft

0.04J 2.5133 10

32 32

A

B

m

dm

dm

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9 7

1

9 7

2

Torsional stiffness of engine shaft

80 10 6.136 10

0.25Torsional stiffness of impeller shaft

80 10 2.5133 10

0.15Mass moment of inertia of equ

A

B

GJ

L

GJ

L

196

1

' 2

ivalent impeller

Also, Torsional stiffness of equivalent impeller shaft

BI

15 kgm

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t eq

)

Equivalent system

500 kgm2

15 kgm2

K =28625.3 Nm/radt

L=1.7148m

0 As it is a semi definite system

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1

1 22

1 2

2

1 2

0

k (I ) 28625.3 (500 15)Also = = 44.34 / sec

I 500 15

Location of node on the equivalent shaft from left side rotor

15x = 1.7148

500 15

t I radI

IL

I I

As it is a semi definite system,

0.0499 0.05 .m

1 1

-1

Node


xL

/I2 I1

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Systems with two degree of freedom

Forced Vibration

m1

K1

m2

K2

x1

x2

F1 tsinFF 01

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Problem-1Governing equations

Newtons method

. Force equilibrium diagram

m1

K1x1

K2(x2-x1)

11xm

m2 22xm

m1

K1

m2

K2

x1

x2

F1

tsinFF 01

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Newtons method

. From Force equilibrium diagram of massm1

tsinF)x(xKxKxm 01221111

tsinFxK)xK(Kxm 02212111 1st Eqn.

of motion

. From Force equilibrium diagram of massm2

0)x(xKxm 12222

0xKxKxm 221222 2nd Eqn. of motion

Governing equations

m1

K1x1

K2(x2-x1)

11xm

m2 22xm

tsinFF

0

1

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Solution of governing equationsIt is possible to have pure harmonic free vibration for both themasses.Let us assume )t 1x Asin(

2x Bsin(t + )

Be the forced response of the system, where A and B are theamplitudes to be obtained

The above equations have to satisfy the governing equations

of motions

0xKxKxm 221222

1st Eqn. of motion

2nd Eqn. of motion

tsinFxK)xK(Kxm 02212111

where is the forcing

frequency

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Solution of governing equations

2

1 2 1 2 0(K K ) m Asin(t ) K Bsin(t ) F sint

22 2 2K Asin(t ) (K m )Bsin(t ) 0

0)sin( tIn above equations

B 2

1 2 1 2 0(K K ) m A K F sint

22 2 2K A (K m )B 0

The above equations reduces to: (characteristic equation)

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B 2

1 2 1 2 0(K K ) m A K F sint

22 2 2K A (K m )B 0

From above equations obtain A1 and A2 by Cramers rule


20

22 2

KF........

0 K m A

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B

201 2 1

2

F(K K ) m

0K

0

mK

K........

K

m)K(K

2

22

2

2

2

121

where is the determinant of characteristic equations


222222121 KmKm)K(K

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m1

K1

m2

K2

x1

x2

F1


if is the natural frequency of system-I

1

11

m

K

if is the natural frequency of system-II

2

22

m

K

We have two resonant frequencies

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Dynamic vibration absorber

m2

K2

x2

x1

m1

K1

F1

Main system

having forcedvibration

Absorbersystem

The system can be used asDynamic vibration absorber

(Tuned damper)

Choose K2, m2 such thatvibrations of mass m1 isminimized

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If the system has to be used as Dynamic vibration absorber,the amplitude of vibration of mass m1, i.e. A=0

20

2

2 2

KF0 K m

A 0

20

2

2 2

KF0

0 K m

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0mKF 2220 )(


20

2

2 2

KF0

0 K m

0mK 222

2

22

m

K

2

2

mK

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2

2

m

K


Natural frequency of absorber system

2

2

2 m

K Excitation frequency of the system isequal to natural frequency of absorber

system

This shows that if excitation frequency of the system isequal to natural frequency of absorber system, thenvibration of the main system is zero

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Undesired maximum vibrations can occur only when mainsystem is under resonance, or near it i.e. 1

So, to reduce undesired vibration of main system mass m1,we choose K2, m2 in such a way that 2 1

2

2

1

1

m

K

m

K

2

2

1

1

m

K

m

KOR


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2

0 2 2

2 2 2

1 2 1 2 2 2

F (K m )A

(K K ) m K m K

20

2

2 2

KF

0 K m A


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y

To find the natural frequencies of the entire system, equatethe denominator of A to zero.

0mK

K........

K

m)K(K

2

22

2

2

2

121

1 2

2 2

2 1 2 2

1 1 2 1

k yields,

1 1 0k m m k k k k k

2 2 21 2 1 2 2 2 (K K ) m K m K 0

Dividing the entire equation by k

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Problem-1

Dynamic vibration Absorber

A reciprocating machine of mass M runs at aconstant speed of N rpm. After it was

installed, it was found that the forcingfrequency is too close to the naturalfrequency of the machine. What dynamicabsorber should be added, if the nearestnatural frequency of the system should be atleast 25% from the impressed frequency.

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Problem-1


Machine

Vibration absorber

Data Given

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Data Given

Natural frequencies of machine must be 25% away fromthe operating speed.Hence r=0.75

4 2

4 2

2 2

1 1

1 1

2 2

we have;

(2 ) 1 0

(0.75) (2 )(0.75) 1 0

0.34028. And =

where m , are mass and stiffness of the main systemand m , are mass and stiffness of the absorber system

Mass of abso

r r

m k

m k

kk

Hence

1

1

rber =0.34028 m 0.34028

Stiffness of absorber =0.34028k

M

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Problem-2

A machine part vibrates with large amplitudeswhen the compressor speed is 250 rpm. Tostudy the vibration, a springmass system is

suspended from the machine part to act as anabsorber. A 2kg absorber mass tuned to 250cpm resulted in two resonant frequencies of200 cpm and 300 cpm. Determine the

equivalent spring and mass constants for themachine part.

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Date post:	05-Apr-2018
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